This Exam P sample reference tests Uniform Distribution. Writing the area as pi times the squared radius reduces the calculation to the second and fourth moments of a uniform radius. Those moments give variance 36pi²⁄⁵, so choice D is correct.
These notes identify the calculation error associated with each wrong letter when that error is reproducible.
AThis squares Var(R)=3/4 and multiplies by pi², giving pi²(3/4)²=9pi²⁄¹⁶. Variance does not commute with squaring the variable.
BThis applies the linear scaling rule as though area were pi R, giving pi² Var(R)=3pi²⁄⁴ and omitting the squared-radius transformation.
CThis doubles the preceding linearized result merely because the area contains R², producing 2pi² Var(R)=3pi²⁄². The nonlinear moment calculation is still required.
EThis reports E[A²]=pi² E[R⁴]=81pi²⁄⁵ without subtracting (E[A])²=9pi².
Original practice · fully worked
Original variant: energy variability from triangular voltage
A calibration voltage V has density f(v)=v/8 for 0<v<4. A device records the energy index W=5V². Calculate Var(W).
A 200/9
B 1600/3
C 5120/9
D 6400/3
E 6400
Variant answer in brief
The triangular density gives E[V²]=8 and E[V⁴]=256/3. Thus Var(V²)=64/3, and scaling by 5 gives Var(W)=25(64/3)=1600/3, so choice B is correct.
Setup
Setup
Check the supplied density and write the two moments needed for the squared transformation.
∫048vdv=1
Var(W)=25Var(V2)
Model
Model
Evaluate the second and fourth raw moments under the triangular density.
E[V2]=∫04v28vdv=8
E[V4]=∫04v48vdv=3256
Compute
Compute
Center the fourth moment and apply the squared scale factor.
The 3108-page Probability Proof Manual reorganizes 718 verified Exam P solutions by syllabus skill and adds formula proofs, error patterns, and original worked practice.